Question

cho hai số a,b thỏa mãn a.b lớn hơn bằng 1.chứng minh rằng:\(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\)lớn hơn bằng\(\dfrac{2}{1+ab}\)

Answer 1

\(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}=\dfrac{a^2+b^2+2}{a^2b^2+a^2+b^2+1}=1-\dfrac{a^2b^2-1}{a^2b^2+a^2+b^2+1}\ge1-\dfrac{a^2b^2-1}{a^2b^2+2ab+1}\)

\(=1-\dfrac{\left(ab-1\right)\left(ab+1\right)}{\left(ab+1\right)^2}=1-\dfrac{ab-1}{ab+1}=\dfrac{2}{ab+1}\) (đpcm)

Dấu "=" xảy ra khi \(a=b\)

Answer 2

\(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\)

\(\Rightarrow\left(\dfrac{1}{1+a^2}-\dfrac{1}{1+ab}\right)+\left(\dfrac{1}{1+b^2}-\dfrac{1}{1+ab}\right)\ge0\)

\(\Rightarrow\dfrac{ab-a^2}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{ab-b^2}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Rightarrow\dfrac{a\left(b-a\right)}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{b\left(a-b\right)}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Rightarrow\dfrac{a\left(b-a\right)\left(1+b^2\right)+b\left(a-b\right)\left(1+a^2\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Rightarrow\dfrac{\left(b-a\right)\left(a+ab^2\right)-\left(b-a\right)\left(b+a^2b\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Rightarrow\dfrac{\left(b-a\right)\left(-\left(b-a\right)+ab\left(b-a\right)\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Rightarrow\dfrac{\left(b-a\right)^2\left(ab-1\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\) (luôn đúng vì \(ab\ge1\))